Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}-8x-4y &= -2 \\ -x-5y &= -7\end{align*}$
Explanation: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-5$ and the bottom equation by $4$ $\begin{align*}40x+20y &= 10\\ -4x-20y &= -28\end{align*}$ Add the top and bottom equations. $36x = -18$ Divide both sides by $36$ and reduce as necessary. $x = -\dfrac{1}{2}$ Substitute $-\dfrac{1}{2}$ for $x$ in the top equation. $-8( -\dfrac{1}{2})-4y = -2$ $4-4y = -2$ $-4y = -6$ $y = \dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{1}{2}, \enspace y = \dfrac{3}{2}$.